I have numbers in one cell that are separated by "-" and I want to multiply them by another number in another cell.
Example:
A: 12 - 6 - 8
B: 83
Doing MULTIPLY(SUM(SPLIT(A,"-")),B) is not mathematically correct. How can I do 12 * 83 + 8 * 83 + 8 * 83?
Thanks a lot.
Here's what you can try:
=INDEX(LAMBDA(ζ,JOIN("-",ζ*SEQUENCE(1,COLUMNS(ζ))))(SPLIT(A1,"-")))
Update
=PRODUCT(SPLIT(A1,"-"))
Update 2
Let's say you have 83 in B1 and 12 - 6 - 8 in A1. Here's what you can try:
=LAMBDA(ζ,SUMPRODUCT(ζ,IF(ζ,B1)))(SPLIT(A1,"-"))
Related
I'm relatively new to lua and programming in general (self taught), so please be gentle!
Anyway, I wrote a lua script to read a UDP message from a game. The structure of the message is:
DATAxXXXXaaaaBBBBccccDDDDeeeeFFFFggggHHHH
DATAx = 4 letter ID and x = control character
XXXX = integer shows the group of the data (groups are known)
aaaa...HHHHH = 8 single-precision floating point numbers
The last ones is those numbers I need to decode.
If I print the message as received, it's something like:
DATA*{V???A?A?...etc.
Using string.byte(), I'm getting a stream of bytes like this (I have "formatted" the bytes to reflect the structure above.
68 65 84 65/42/20 0 0 0/237 222 28 66/189 59 182 65/107 42 41 65/33 173 79 63/0 0 128 63/146 41 41 65/0 0 30 66/0 0 184 65
The first 5 bytes are of course the DATA*. The next 4 are the 20th group of data. The next bytes, the ones I need to decode, and are equal to those values:
237 222 28 66 = 39.218
189 59 182 65 = 22.779
107 42 41 65 = 10.573
33 173 79 63 = 0.8114
0 0 128 63 = 1.0000
146 41 41 65 = 10.573
0 0 30 66 = 39.500
0 0 184 65 = 23.000
I've found C# code that does the decode with BitConverter.ToSingle(), but I haven't found any like this for Lua.
Any idea?
What Lua version do you have?
This code works in Lua 5.3
local str = "DATA*\20\0\0\0\237\222\28\66\189\59\182\65..."
-- Read two float values starting from position 10 in the string
print(string.unpack("<ff", str, 10)) --> 39.217700958252 22.779169082642 18
-- 18 (third returned value) is the next position in the string
For Lua 5.1 you have to write special function (or steal it from François Perrad's git repo )
local function binary_to_float(str, pos)
local b1, b2, b3, b4 = str:byte(pos, pos+3)
local sign = b4 > 0x7F and -1 or 1
local expo = (b4 % 0x80) * 2 + math.floor(b3 / 0x80)
local mant = ((b3 % 0x80) * 0x100 + b2) * 0x100 + b1
local n
if mant + expo == 0 then
n = sign * 0.0
elseif expo == 0xFF then
n = (mant == 0 and sign or 0) / 0
else
n = sign * (1 + mant / 0x800000) * 2.0^(expo - 0x7F)
end
return n
end
local str = "DATA*\20\0\0\0\237\222\28\66\189\59\182\65..."
print(binary_to_float(str, 10)) --> 39.217700958252
print(binary_to_float(str, 14)) --> 22.779169082642
It’s little-endian byte-order of IEEE-754 single-precision binary:
E.g., 0 0 128 63 is:
00111111 10000000 00000000 00000000
(63) (128) (0) (0)
Why that equals 1 requires that you understand the very basics of IEEE-754 representation, namely its use of an exponent and mantissa. See here to start.
See #Egor‘s answer above for how to use string.unpack() in Lua 5.3 and one possible implementation you could use in earlier versions.
I have been working on this program for hours and cannot find out how to make the numbers loop around after they hit saturday. They either go way passed it to the right or if i add and endl; they go up and down.
// This is how my output looks like (except they curve around they just go forever to the right:
Number of days: 31
Offset: 0
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
This what you mean?
#include <iostream>
using namespace std;
int main()
{
int i;
for (i=1; i<=31; i++) {
cout << ((i<10) ? " " : "") << i << " ";
if (i%7==0) cout << endl;
}
return 0;
}
Outputs:
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31
The % sign is the modulus operator. It gives the remainder of division. So every 7th day divided by 7 is going to have a remainder of zero. That's how you check where to put the line breaks.
Using the Images package, I can open up a color image, convert it to Gray scale and then :
using Images
img_gld = imread("...path to some color jpg...")
img_gld_gs = convert(Image{Gray},img_gld)
#change from floats to Array of values between 0 and 255:
img_gld_gs = reinterpret(Uint8,data(img_gld_gs))
Now I've got a 1920X1080 array of Uint8's:
julia> img_gld_gs
1920x1080 Array{Uint8,2}
Now I want to get a histogram of the 2D array of Uint8 values:
julia> hist(img_gld_gs)
(0.0:50.0:300.0,
6x1080 Array{Int64,2}:
1302 1288 1293 1302 1297 1300 1257 1234 … 12 13 13 12 13 15 14
618 632 627 618 623 620 663 686 189 187 187 188 185 183 183
0 0 0 0 0 0 0 0 9 9 8 7 8 7 7
0 0 0 0 0 0 0 0 10 12 9 7 13 7 9
0 0 0 0 0 0 0 0 1238 1230 1236 1235 1230 1240 1234
0 0 0 0 0 0 0 0 … 462 469 467 471 471 468 473)
But, instead of 6x1080, I'd like 256 slots in the histogram to show total number of times each value has appeared. I tried:
julia> hist(img_gld_gs,256)
But that gives:
(2.0:1.0:252.0,
250x1080 Array{Int64,2}:
So instead of a 256x1080 Array, it's 250x1080. Is there any way to force it to have 256 bins (without resorting to writing my own hist function)? I want to be able to compare different images and I want the histogram for each image to have the same number of bins.
Assuming you want a histogram for the entire image (rather than one per row), you might want
hist(vec(img_gld_gs), -1:255)
which first converts the image to a 1-dimensional vector. (You can also use img_gld_gs[:], but that copies the data.)
Also note the range here: the hist function uses a left-open interval, so it will omit counting zeros unless you use something smaller than 0.
hist also accepts a vector (or range) as an optional argument that specifies the edge boundaries, so
hist(img_gld_gs, 0:256)
should work.
I tried to create a neural network to estimate y = x ^ 2. So I created a fitting neural network and gave it some samples for input and output. I tried to build this network in C++. But the result is different than I expected.
With the following inputs:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 -1
-2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15 -16 -17 -18 -19 -20 -21 -22 -23 -24 -25 -26 -27 -28 -29 -30 -31 -32 -33 -34 -35 -36 -37 -38 -39 -40 -41 -42 -43 -44 -45 -46 -47 -48 -49 -50 -51 -52 -53 -54 -55 -56 -57 -58 -59 -60 -61 -62 -63 -64 -65 -66 -67 -68 -69 -70 -71
and the following outputs:
0 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400
441 484 529 576 625 676 729 784 841 900 961 1024 1089 1156 1225 1296
1369 1444 1521 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401 2500
2601 2704 2809 2916 3025 3136 3249 3364 3481 3600 3721 3844 3969 4096
4225 4356 4489 4624 4761 4900 5041 1 4 9 16 25 36 49 64 81 100 121 144
169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841
900 961 1024 1089 1156 1225 1296 1369 1444 1521 1600 1681 1764 1849
1936 2025 2116 2209 2304 2401 2500 2601 2704 2809 2916 3025 3136 3249
3364 3481 3600 3721 3844 3969 4096 4225 4356 4489 4624 4761 4900 5041
I used fitting tool network. with matrix rows. Training is 70%, validation is 15% and testing is 15% as well. The number of hidden neurons is two. Then in command lines I wrote this:
purelin(net.LW{2}*tansig(net.IW{1}*inputTest+net.b{1})+net.b{2})
Other information :
My net.b[1] is: -1.16610230053776 1.16667147712026
My net.b[2] is: 51.3266249426358
And net.IW(1) is: 0.344272596370387 0.344111217766824
net.LW(2) is: 31.7635369693519 -31.8082184881063
When my inputTest is 3, the result of this command is 16, while it should be about 9. Have I made an error somewhere?
I found the Stack Overflow post Neural network in MATLAB that contains a problem like my problem, but there is a little difference, and the differences is in that problem the ranges of input and output are same, but in my problem is no. That solution says I need to scale out the results, but how can I scale out my result?
You are right about scaling. As was mentioned in the linked answer, the neural network by default scales the input and output to the range [-1,1]. This can be seen in the network processing functions configuration:
>> net = fitnet(2);
>> net.inputs{1}.processFcns
ans =
'removeconstantrows' 'mapminmax'
>> net.outputs{2}.processFcns
ans =
'removeconstantrows' 'mapminmax'
The second preprocessing function applied to both input/output is mapminmax with the following parameters:
>> net.inputs{1}.processParams{2}
ans =
ymin: -1
ymax: 1
>> net.outputs{2}.processParams{2}
ans =
ymin: -1
ymax: 1
to map both into the range [-1,1] (prior to training).
This means that the trained network expects input values in this range, and outputs values also in the same range. If you want to manually feed input to the network, and compute the output yourself, you have to scale the data at input, and reverse the mapping at the output.
One last thing to remember is that each time you train the ANN, you will get different weights. If you want reproducible results, you need to fix the state of the random number generator (initialize it with the same seed each time). Read the documentation on functions like rng and RandStream.
You also have to pay attention that if you are dividing the data into training/testing/validation sets, you must use the same split each time (probably also affected by the randomness aspect I mentioned).
Here is an example to illustrate the idea (adapted from another post of mine):
%%# data
x = linspace(-71,71,200); %# 1D input
y_model = x.^2; %# model
y = y_model + 10*randn(size(x)).*x; %# add some noise
%%# create ANN, train, simulate
net = fitnet(2); %# one hidden layer with 2 nodes
net.divideFcn = 'dividerand';
net.trainParam.epochs = 50;
net = train(net,x,y);
y_hat = net(x);
%%# plot
plot(x, y, 'b.'), hold on
plot(x, x.^2, 'Color','g', 'LineWidth',2)
plot(x, y_hat, 'Color','r', 'LineWidth',2)
legend({'data (noisy)','model (x^2)','fitted'})
hold off, grid on
%%# manually simulate network
%# map input to [-1,1] range
[~,inMap] = mapminmax(x, -1, 1);
in = mapminmax('apply', x, inMap);
%# propagate values to get output (scaled to [-1,1])
hid = tansig( bsxfun(#plus, net.IW{1}*in, net.b{1}) ); %# hidden layer
outLayerOut = purelin( net.LW{2}*hid + net.b{2} ); %# output layer
%# reverse mapping from [-1,1] to original data scale
[~,outMap] = mapminmax(y, -1, 1);
out = mapminmax('reverse', outLayerOut, outMap);
%# compare against MATLAB output
max( abs(out - y_hat) ) %# this should be zero (or in the order of `eps`)
I opted to use the mapminmax function, but you could have done that manually as well. The formula is a pretty simply linear mapping:
y = (ymax-ymin)*(x-xmin)/(xmax-xmin) + ymin;
I have a mathematical equation and How can I find the it's reverse ?
My equation:
var
x,y:integer;
begin
//example x=1234;
x-(x div 100):=y
end;
after the code I konw "y" how can I find the "x"?(1234)
In general, you can't. Since div does integer division, there are potentially many inputs that can/will produce the same result. Starting from that result, and of those inputs is an equally likely possibility as the original input. For example:
175 div 7 = 25
176 div 7 = 25
177 div 7 = 25
178 div 7 = 25
179 div 7 = 25
180 div 7 = 25
181 div 7 = 25
Starting from 25, any of those numbers from 175 to 181 would be an equally viable answer.